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Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2. We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$. For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$.

Step 3: Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here.

We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality $$ tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1} $$ to prove. Now, note that, according to Step 2, $$ \begin{aligned} &\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \\ &=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1} \end{aligned} $$ Multiplying by $t^{k+1}$ and adding up, we get $$ t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1} $$ which is exactly what we need.

The end.

Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2. We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$. For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$.

Step 3: Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here.

We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality $$ tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1} $$ to prove. Now, note that, according to Step 2, $$ \begin{aligned} &\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \\ &=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1} \end{aligned} $$ Multiplying by $t^{k+1}$ and adding up, we get $$ t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1} $$ which is exactly what we need.

The end.

P.S. If somebody is still interested, the bottom line is almost trivial once the top line is known. Assume again that $a>b$, $a+b=1$. Put $t=a-b$.

$$ \begin{aligned} &\left(\frac{a^b}{2^b}+\frac{b^a}{2^a}\right)^2=(a^{2b}+b^{2a})(2^{-2b}+2^{-2a})-\left(\frac{a^b}{2^a}-\frac{b^a}{2^b}\right)^2 \\ &\le 1+\frac 14\{ [\sqrt 2(2^{t/2}-2^{-t/2})]^2-[(1+t)^b-(1-t)^a]^2\} \end{aligned} $$ Now it remains to note that $2^{t/2}-2^{-t/2}$ is convex on $[0,1]$, so, interpolating between the endpoints, we get $\sqrt 2(2^{t/2}-2^{-t/2})\le t$. Also, the function $x\mapsto (1+x)^b-(1-x)^a$ is convex on $[0,1]$ (the second derivative is $ab[(1-x)^{b-2}-(1+x)^{a-2}]$, which is clearly non-negative). But the derivative at $0$ is $a+b=1$, so $(1+x)^b-(1-x)^a\ge x$ on $[0,1]$. Plugging in $x=t$ finishes the story.

OOPS, Step 2 is wrong so far (the constant $c$ depends on $t$!). I'll see if I can fix it.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2. Recall that for $0<t<1$, one has $\lambda^t=c\int_0^{\infty}z^{-t}(1-e^{-\lambda z})\frac{dz}z$ with some positive $c$. Thus $$ (1-t)^t-(a-t)^t=c\int_0^\infty \left(\frac {e^z}z\right)^t(e^{-az}-e^{-z})\frac{dz}z $$ Since for $Q>1$ the Taylor coefficients of $Q^t$ are all positive, we conclude that the Taylor coefficients of $(1-t)^t-(a-t)^t$ are all positive. In particular, the difference is not less than its second order Taylor polynomial on $(0,a)$, which is $$ -t^2+t\log\frac 1{a}+t^2\left[\frac 1a-\frac{\log^2\frac 1a}2\right]\ge -t^2+t\log\frac 1a+t^2\left[1+\log\frac 1a\right] $$

Step 3: We thus have $$ \begin{aligned} b^{2a}&=b(a-t)^{t}\le b\left[(1-t)^t+t^2-t^2-t(1+t)\log\frac 1a\right] \\ &\le b\left[1-t^2-t(1+t)\log\frac 1a\right] \end{aligned} $$ because $(1-t)^t\le 1-t^2$ by Bernoulli.

It remains to add the results of Step 1 and Step 3.

Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2. We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$. For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$.

Step 3: Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here.

We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality $$ tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1} $$ to prove. Now, note that, according to Step 2, $$ \begin{aligned} &\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \\ &=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1} \end{aligned} $$ Multiplying by $t^{k+1}$ and adding up, we get $$ t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1} $$ which is exactly what we need.

The end.

Another typical AoPS question. The inequality is not complicated but requires being minimally comfortable with Taylor expansions of elementary functions. The proof consists of 3 straightforward steps.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2. Recall that for $0<t<1$, one has $\lambda^t=c\int_0^{\infty}z^{-t}(1-e^{-\lambda z})\frac{dz}z$ with some positive $c$. Thus $$ (1-t)^t-(a-t)^t=c\int_0^\infty \left(\frac {e^z}z\right)^t(e^{-az}-e^{-z})\frac{dz}z $$ Since for $Q>1$ the Taylor coefficients of $Q^t$ are all positive, we conclude that the Taylor coefficients of $(1-t)^t-(a-t)^t$ are all positive. In particular, the difference is not less than its second order Taylor polynomial on $(0,a)$, which is $$ -t^2+t\log\frac 1{a}+t^2\left[\frac 1a-\frac{\log^2\frac 1a}2\right]\ge -t^2+t\log\frac 1a+t^2\left[1+\log\frac 1a\right] $$

Step 3: We thus have $$ \begin{aligned} b^{2a}&=b(a-t)^{t}\le b\left[(1-t)^t+t^2-t^2-t(1+t)\log\frac 1a\right] \\ &\le b\left[1-t^2-t(1+t)\log\frac 1a\right] \end{aligned} $$ because $(1-t)^t\le 1-t^2$ by Bernoulli.

It remains to add the results of Step 1 and Step 3.

OOPS, Step 2 is wrong so far (the constant $c$ depends on $t$!). I'll see if I can fix it.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2. Recall that for $0<t<1$, one has $\lambda^t=c\int_0^{\infty}z^{-t}(1-e^{-\lambda z})\frac{dz}z$ with some positive $c$. Thus $$ (1-t)^t-(a-t)^t=c\int_0^\infty \left(\frac {e^z}z\right)^t(e^{-az}-e^{-z})\frac{dz}z $$ Since for $Q>1$ the Taylor coefficients of $Q^t$ are all positive, we conclude that the Taylor coefficients of $(1-t)^t-(a-t)^t$ are all positive. In particular, the difference is not less than its second order Taylor polynomial on $(0,a)$, which is $$ -t^2+t\log\frac 1{a}+t^2\left[\frac 1a-\frac{\log^2\frac 1a}2\right]\ge -t^2+t\log\frac 1a+t^2\left[1+\log\frac 1a\right] $$

Step 3: We thus have $$ \begin{aligned} b^{2a}&=b(a-t)^{t}\le b\left[(1-t)^t+t^2-t^2-t(1+t)\log\frac 1a\right] \\ &\le b\left[1-t^2-t(1+t)\log\frac 1a\right] \end{aligned} $$ because $(1-t)^t\le 1-t^2$ by Bernoulli.

It remains to add the results of Step 1 and Step 3.